+2 UNIT 4 PAGE- 3
SPONTANEOUS PROCESSES
Processes which takes place by itself under a given set of conditions without the intervention of any kind are known as spontaneous processes. All the processes that take place in nature are spontaneous in character, proceed only in one direction and are therefore thermodynamically irreversible. They can be reversed only by the help of an external agency. Let us consider some examples of spontaneous changes to understand their irreversible nature and the conditions which determine their feasibility.
i) We find that a beaker containing hot water kept in room cools spontaneously by losing heat to the surroundings. The first law of thermodynamics is obeyed because the energy lost by the system (i.e., beaker containing water) is gained by the surroundings. However, the first law will still be obeyed in the reverse process, namely, when the surrounding lose energy and the system gains it. We never observe such situation where the beaker of water at room temperature becomes hot on its own. In short , the cooling of hot body in a room proceeds spontaneously but not its heating even though both processes are allowed by the first law.
ii) If we add a drop of ink in a beaker containing water , the ink spreads throughout the beaker making the whole solution coloured. The reverse process where dispersed ink spontaneously accumulates to form a single drop does not take place.
iii) Two gases under the same pressure in two identical bulbs connected through a stopcock , mix and give a homogeneous mixture on opening the stopcock.
Spontaneous mixing of two gases
(a) before mixing (b) after mixing
Consider few examples involving chemical changes.
(i) If equal volumes of equimolar solutions of hydrochloric acid and sodium hydroxide are mixed, the neutralisation reaction proceeds spontaneously yielding sodium chloride and water. However, if sodium chloride is dissolved in water, the reverse reaction does not take place on its own.
In general, all transformations under given experimental conditions have a characteristic direction in which they proceed spontaneously ; i.e., on their own with out intervention of an outside agency. The reverse process i.e., the beaker of water can be made hot by heating, but this is not a spontaneous process because an outside agency (a gas burner) has to be used. Similarly an ink solution can be processed to obtain the ink drop ; and two gases in a mixture can be separated if allowed to diffuse through a porous plug, and a solution of sodium chloride can be made to yield hydrochloric acid and sodium hydroxide. In each case, the change has to be brought about using some external agency. It does not take place on its own.
There is another feature of all spontaneous changes that may be noted. A beaker of hot water cools until it attains the temperature of the surroundings after which no further change in temperature is observed. Now the beaker of water and the surroundings are in thermal equilibrium. The drop of ink similarly separates until it is distributed uniformly. Mixing of two gases continues until each is evenly distributed throughout both the containers. Thus all spontaneous changes proceed till equilibrium is achieved.
Energy and Spontaneity
Common experience shows that spontaneous processes involving macroscopic objects proceed with a decrease in potential energy.
(i) A rock spontaneously rolls down a slope if it is dislodged, but never rolls up to the top again (Fig)
The rolling of a rock down hill The rolling of a rock uphill
is a spontaneous process. is a non-spontaneous process.
The rock eventually comes to
Equilibrium at the bottom of the
hill.
(ii) If we drop a bottle , it falls spontaneously to the ground.
(iii) When octane is ignited , it burns spontaneously in oxygen to give CO2 and H2O according to the reaction :
2 C8H18 + 25 O2(g) ® 16 CO2(g) + 18 H2O(ℓ)
: ∆Hө = - 10951 kJ/mol
Heat is evolved and the reaction is exothermic. Thus the system moves from a position of higher to lower energy.
But there are many reactions which are spontaneous for which ∆Hө is +ve or zero.
(iv) The evaporation of liquid water from an open vessel is spontaneous, but it is a process in which the system requires energy to evaporate water molecules.
(v) Dissolution of a solute in a solvent is spontaneous eventhough the process may be endothermic, i.e., it may require energy.
(vi) Dinitrogen pentoxide (N2O5) decomposes spontaneously at room temperature into NO2 and O2 although the reaction is endothermic.
2 N2O5(g) ® 4 NO2 (g) + O2 (g) : ∆Hө = 219.0 kJ/mol
(vii) Expansion of an ideal gas through a pinhole is a spontaneous process. In this expansion there is no energy change of the system.
Experience tells us that highly exothermic processes are generally spontaneous but some endothermic process may also be spontaneous showing that lowering of energy is not the only determining factor for spontaneity.
Disorder, Entropy and Second law of Thermodynamics
During the spontaneous evaporation of water, there is an increase in energy of the molecules and the process is endothermic. However, this leads to greater disorder amongst water molecules in the vapour states compared to the liquid state. The disorder is the manifestation of another thermodynamic property called entropy and is represented by the symbol S. Entropy like any other thermodynamic property such as internal energy (U) and enthalpy(H) should be a state function. The entropy change , DS is given by the equation :
…………..(1)Here qrev is the heat absorbed when the process is carried out reversibly and isothermally (i.e., at constant T). If q is expressed in Joule(J) and temperature in Kelvin (K), then entropy change is given in unit of J/K or J K-1. The DS defined by equation (1) is independent of path , as it is a state function.
qrev / T is is a measure of increase in disorder of the system. This can be justified on the grounds that the absorption of heat into a system leads to the vigorous movement of the molecules due to increased kinetic energy. In other words, the more heat the system absorbs, the more disordered it becomes. If heat is absorbed at low temperature it becomes more disordered than when the same amount of heat is added at higher temperature. This is obvious as T is placed in denomenator in equation 1.
Physical meaning of Entropy
The physical meaning of entropy is that entropy is a measure of disorder (or randomness) of a system. The relation between entropy and disorder provides a suitable explanation for entropy change in various processes. The greater the disorder in a system, the higher the entropy. For a given substance the solid state is the state of lowest entropy (most ordered state) , the gaseous state is the state of highest entropy and liquid state is intermediate between the two.
Thus, when a sample changes from solid to liquid to gas, the particles become increasingly disordered and therefore its entropy increases. This shown in Fig (a)
(a)
(b)
Increase in disorder or entropy when a solid changes to liquid to gas.
Fig (a) shows the increase in disorder when solid changes to liquid to gas. When the solid is heated its entropy increases as its temperature is raised. The entropy increases sharply when solid melts to form more disordered liquid. It further gradually increases again upto boiling point. A second larger jump in entropy occurs when the liquid changes into vapour at the boiling point. The changes are shown in Fig(b).
Similarly , in the case of mixing of two gases when the stopcock is opened, the gases mix to achieve more randomness or disorder. In this case, there is no exchange of matter or energy between the system and surroundings. The change occurs from ordered state (less entropy) to disordered state (higher entropy) . Thus the change in entropy is positive.
Driving force as the overall tendency for a process
The spontaneous processes occur because of two tendencies.
i) Tendency of a system to acquire a state of minimum energy.
ii) Tendency of a system to acquire a state of maximum randomness.
The overall tendency for a process to take place by itself is called driving force. Regarding these two tendencies , it is very important to note the following points.
i) The two tendencies act independent of each other.
ii) The two tendencies may work in the same or in opposite directions in a process.
It is clear from the discussion that when there is no change in energy, the second tendency is the governing factor i.e., reaction will be spontaneous or non-spontaneous depending upon whether randomness factor favours or opposes. On the other hand, if there is no change in randomness, the energy factor is the controlling factor. However, this is not so simple in actual practice as both tendencies operate simultaneously. In these cases, the significance of magnitude of these tendencies determines whether the process is feasible or not.
Second Law of Thermodynamics
The second law of thermodynamics introduces the concept of entropy and its relation with spontaneous process. In an isolated system, such as mixing of gases , there is no exchange of energy and matter between system and surroundings. But due to increase in randomness, there is an increase in entropy. Thus we can say that, for a spontaneous process in an isolated system, the change in entropy is positive , i.e., DS > 0 . However, if the system is not isolated, we have to take into account the entropy changes of the system and the surroundings. Then total entropy changes (DStotal) will be equal to the sum of the change in entropy of the system and the surroundings (DSsurroundings). i.e.,
DStotal = DS system + DSsurroundings
For a spontaneous process , DStotal must be positive i.e.,
DStotal = DSsystem + DSsurroundings > 0
But the system and surroundings constitute ‘universe’ for thermodynamic point of view so that for spontaneous change
DSuniverse > 0 ……… (2)
The relation (2) (DSuniverse > 0) forms the basis of most useful form of Second Law of Thermodynamics which states that : the entropy of the universe always increases in the course of every spontaneous (natural) change. Thus combining statements of First and Second Laws of thermodynamics, we conclude that : In any natural or spontaneous process, the energy of the universe remains constant but the entropy of the universe always increases.
Entropy change and equilibrium
For a spontaneous process DStotal > 0. The careful study of the spontaneous processes reveals that each of them gradually and ultimately passes into a state of equilibrium. For example, in intermixing of gases, the process of intermixing continues till a uniform mixture is formed. Similarly in the evaporation of water in a closed vessel , soon a state of equilibrium is reached. Although these processes are favoured by increase in the value of DStotal , yet no further increase in entropy occurs once the equilibrium is established. This leads to conclude that :
At equilibrium,
DStotal = 0
Following the above discussion, the entropy criterion can be summed up as follows:
· DStotal > 0 : process is spontaneous
· DStotal = 0 : equilibrium state
· DStotal < 0 : process is non-spontaneous
Let us apply , the entropy criterion to decide the feasibility of the process of conversion of water to ice, at 1 atm pressure,
H2O(ℓ) ® H2O(s)
The entropy changes for different temperatures 272, 273 and 274 K are given below.
Temperature (K) | DSsystem J/K/mol | DSsurroundings J/K/mol | DStotal J / K / mol |
272 | - 21.85 | + 21.93 | + 0.08 |
273 | - 21.99 | + 21.99 | 0.00 |
274 | - 22.13 | +22.05 | - 0.08 |
i) DStotal is positive at 272 K, therefore freezing of liquid water to ice is spontaneous.
H2O(ℓ) ® H2O(s) : spontaneous at 272 K
ii) At 274 K, DStotal is negative, therefore the freezing of water is not spontaneous at 274 K.
iii) At 273 K, DStotal is zero, therefore the process is at equilibrium, i.e., neither freezing nor melting is spontaneous. At this temperature, water and ice are in equilibrium and no net change is observed.
H2O(ℓ) ® H2O(s) : equilibrium at 273 K
Thus, we observe that DStotal is a criterion for spontaneity of a change.
Characteristics of Entropy
The impotant characteristics of entropy are summed up as below:
(i) Entropy is an extensive property. Its value depends upon the amount of the substance present in the system.
(ii) Entropy of the system is a state function. It depends upon the state variables (T, P, V and n) . Thus , the change in entropy is given as :
DS = Sfinal state - Sinitial state
(iii) The change in entropy in going from one state to another is independent of the path .
(iv) The change in entropy for a cyclic process is always zero.
(v) The total entropy change of an non-isolated system is equal to the entropy change of the system and entropy change of the surroundings. The sum is called entropy change of the universe.
DS = DSsystem + DSsurr
(vi) In a reversible process in equilibrium, DSuniverse = 0 and therefore
DSsystem = -DSsurr
(vii) In a irreversible process,
DSuniverse > 0 . This means that there is an increase in entropy of universe in spontaneous changes.
Entropy changes in system and surroundings and total entropy change for Exothermic and Endothermic Reactions
Heat increases the thermal motion of the atoms or molecules and increases their disorder and hence their entropy. In the case of exothermic process, the heat escapes into the surroundings and therefore , entropy of the surroundings increases. On the other hand in the case of endothermic process , the heat enters the system from the surroundings and therefore , the entropy of the surroundings decreases. These changes are shown in Fig.
Entropy changes in (a) Exothermic and (b) Endothermic processes.
In general , there will be an overall increase of the total entropy (or disorder) whenever the disorder of the surroundings is greater than the decrease in disorder of the system. The process will be spontaneous only when the total entropy increases.
Let us consider an exothermic process. In exothermic processes , heat released by the reaction increases the entropy of the surroundings. The overall entropy changes is certainly positive when the entropy of the system is positive (Fig a)
Entropy changes for an exothermic reaction
In some exothermic reactions, entropy of the system may decrease. This is possible in the case of reactions which involve the conversion of a gas into solid product. However, if the reaction is highly exothermic and increase in entropy of the surroundings is very high, the total entropy change will be positive and the reaction will be spontaneous (Fig b).
For example, consider the reaction of oxidation of magnesium to magnesium oxide. The reaction is highly exothermic.
2 Mg(s) + O2(g) ® 2 MgO(s) : DHΘ = - 1202 kJ mol-1
But the reaction proceeds by decrease in entropy and DSΘ for the reaction is -217 J K-1mol-1.
The heat released to the surroundings will result into increase in entropy of the surroundings. The same formula applies to change in entropy of the surroundings that accompanies a chemical reaction.
Since DStotal is is positive , the reaction will be spontaneous.
In case of endothermic reactions, reactants absorb energy and the products are at higher energy state and the temperature of the system falls. In other words, heat flows from surroundings into the system. As a result, the entropy of the surroundings decreases. The reactions will be spontaneous only when entropy of system increases enough to overcome the decrease in entropy of the surroundings so that overall entropy change is positive. This is shown in Fig.
Endothermic reaction is spontaneous only if the entropy of the system increases enough to overcome the decrease in entropy of the surroundings.
Thus,
· A chemical reaction is spontaneous if it is accompanied by an increase in the total entropy of the system and the surroundings.
· Spontaneous exothermic reactions are common because they release heat that increases the entropy of the surroundings.
· Endothermic reactions are spontaneous only if the reaction mixture undergoes a large increase in entropy.
Entropy Changes of Phase transitions
Phase transition refers to changes of one state (solid, liquid or gas) to another. Such change occurs at a definite temperature. For example, conversion of solid into liquid occurs at the melting point, while changes of liquid into vapour occurs at the boiling point. The entropy change taking place during such transitions can be calculated as follows:
1. Entropy of fusion : Entropy of fusion may be defined as the entropy change taking place when one mole of substance changes from solid state into liquid state at its melting point.
Solid ⇌ Liquid
During this phase transition, the system absorbs heat at constant temperature and pressure. The heat absorbed is equal to the enthalpy of fusion (DHfusion). The entropy change taking place during this transition can be calculated as :
Here, DSfusion = Entropy of fusion
DHfusion = Enthalpy of fusion
Sℓ= Entropy of one mole of liquid at its
melting point
Ss = Entropy of one mole of solid substance
at its melting point.
Tf = Melting point of the substance.
Since DHfusion is positive , therefore DSfusion is also positive. Hence Sℓ > Ss i.e., entropy of liquid substance is greater than the entropy of the substance in the solid state.
2. Entropy of vaporisation : Entropy of vaporisation may be defined as the entropy change taking place when one mole of the substance changes from liquid state into vapour at its boiling point.
Liquid ⇌ Gas
During this transition, the system absorbs heat at constant temperature and pressure. The heat absorbed is equal to the enthalpy of vaporisation(DHvap) The entropy change for the transition may be calculated as :
Here, DSvap = Entropy of vaporisation
DHvap = Enthalpy of vaporisation
Sℓ = Entropy of one mole of substance in liquid
state at its boiling point.
Sg = Entropy of one mole of substance in the gaseous
state.
Tb = Boiling point of the substance
Since DHvap is positive, therefore DSvap is also positive.
Hence Sg > Sℓ i.e., entropy of substance in gaseous state is greater than the entropy of substance in liquid state.
3. Entropy of sublimation : Entropy of sublimation is
the entropy change when one mole of the solid changes into vapour at a particular temperature T. Thus,
where DSsub = DSvap + DSfusion
DSsub = molar entropy of sublimation at temperature T in Kelvin.
Problems
51. State whether each of the following processes will increase or decrease total energy content of the system :
(a) Heat transferred to surroundings.
(b) Work done by the system
(c) Work done on the system
52. Place the following systems in the order of increasing randomness :
(a) 1 mol of gas X
(b) 1 mol of solid X
(c) 1 mol of liquid X
53. Which of the following processes are accompanied by an increase of entropy :
(a) dissolution of iodine in a solvent
(b) HCl is added to AgNO3 solution and precipitate is obtained.
(c) A partition is removed to allow two gases to mix.
54. Predict the sign of entropy change for each of the following changes of state :
(a) Hg (ℓ) ® Hg(g)
(b) AgNO3(s) ® AgNO3(aq)
(c) I2(g) ® I2(s)
(d) C(graphite) ® C(diamond)
44. Predict which of the following , entropy increases/decreases :
(i) A liquid crystallizes into a solid.
(ii) Temperature of a crystalline solid is raised from 0 K to
115 K.
(iii) 2 NaHCO3(s) ® Na2CO3 + CO2(g) + H2O(g)
(iv) H2(g) ® 2 H(g) (N 177 P 6.9)
55. The enthalpy of vaporisation of liquid diethyl ether is 26 kJ/mol at its boiling point(35°C). Calculate DSΘ for the conversion of :
(a) liquid to vapour, and
(b) vapour to liquid at 35°C.
56. The enthalpy change for the transition of liquid water to steam, DHvap is 40.8 kJ/mol at 373 K . Calculate the entropy change for the process.
57. At 00C ice and water are in equilibrium and DH = 6 kJ mol-1. For the process:
H2O(s) ® H2O(ℓ)
Calculate DS for conversion of ice to liquid water.
58. Calculate the entropy change of 1 g of ice to water at 273 K and one atmospheric pressure. The enthalpy of fusion of ice is 6.025 kJ/mol.
59. Calculate the enthalpy of vaporisation per mol for ethanol. Given DS = 109.8 J K-1mol-1 and boiling point of ethanol = 78.5°C.
60. The enthalpy of vapourisation of benzene is 30.8 kJ mol-1 at its boiling point(80.1°C). Calculate the entropy change in going from (i) liquid to vapour and (ii) vapour to liquid at 80.1°C.
61. A reaction A + B ® C + D + q is found to have a positive entropy change. The reaction will be : (N 182 Ex 6.6)
(i) possible at high temperature.
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Predicting the sign of DS
The process in which entropy of product is greater than the entropy of the reactants have positive value of DS. Such a process is accompanied by increase in randomness.
In many reactions, it becomes possible to predict the sign of DS. Such processes are accompanied by increase in randomness. In many reactions, it becomes possible to predict the sign of DS by simply observing the reactants and products. If the products are more random than the reactants DS is positive and if the products are less random than the reactants, DS is negative.
Now, let us consider some reactions and predict the sign of DS.
i) C(s) + O2(g) ® CO2(g)
In this reaction, the product is more random than the reactants because the product molecules are in gaseous state , whereas one of the reactants is in the solid state. DS would be positive for this reaction.
ii) 2 SO3(g) ® 2 SO2(g) + O2(g)
In this reaction also products are more random than the reactants because more molecules are present in gaseous state in product than in reactants . So DS is +Ve.
iii) NaCl(s) + H2O(ℓ) ® Na+(aq) + Cl- (aq)
This process is also accompanied by increase in randomness because Na+ and Cl- ions are free to move more than in solid NaCl. Hence DS is positive for this type of dissolution process.
Gibb’s Energy (G) and Gibb’s Energy Change (DG)
Gibb’s energy of a system is defined as the amount of energy available to a system during a process that can be converted into useful work. In other words, it is a measure of capacity of a system to do useful work denoted by symbol G and is given by :
G = H - T S
where H is the enthalpy of the system, S is the entropy and T is the absolute temperature.
Now, H = U + P V
G = U + PV - T S
The change in Gibb’s energy may be expressed as :
DG = DU + D(PV) - D TS
If the process is carried out at constant temperature and pressure, the terms D(PV) and D(TS) becomes :
D(PV) = P DV and D(TS) = TDS
\ DG = DE + PDV - TDS
But , DU + PDV = DH
or DG = DH - T DS …….. (11)
This equation is called Gibb’s energy equation and is very useful in predicting the spontaneity of a process.
Units of DG
DG has units of energy ie., kJ/mol or J mol-1 because DH and TDS are are energy terms.
DG( J mol-1)= DH( J mol-1) - TDS ( K J K-1 mol-1)
Physical significance of Gibb’s energy
In order to understand the physical significance of Gibb’s energy, let us consider the relationship between heat (q) and work (w). According to first law of thermodynamics :
DU = q + w
Here w includes the pressure-volume work (- PDV ) as well as non-expansion work. Therefore, the above relation can be written as :
DU = q - PDV + wnon-expansion
q = DU + PDV - wnon-expansion
= DH - wnon-expansion ……… (15)
For a change taking place under isothermal and reversible conditions, the entropy change DS is given by :
DS = (q / T)
or q = T DS …….. (16)
Substituting this value of q in Equation (15) :
TDS = DH - wnon-expansion
DH - TDS = wnon-expansion
DGT,P = wnon-expansion
- DGT,P = - wnon-expansion
Decrease in Gibb’s energy = Non-expansion work done by the system
The non-expansion work may be regarded as useful work. Thus, the decrease in Gibb’s energy of a system during a process is a measure of maximum useful work done during the change. Therefore Gibb’s energy G of a system during a process is a measure of the maximum useful work done during the change. Therefore Gibb’s energy , G of a system is a measure of its capacity of doing useful work. Therefore greater the Gibb’s energy change , the greater is the amount of work that can be obtained from the process.
In our bodies , the food stuffs are oxidised to obtain energy to do work. For example, consider the metabolism of glucose in our body. When one mole of glucose (180.2 g) is oxidised
C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(ℓ)
: DHΘ = -2808 kJ/mol : DGΘ = -2870 kJ/mol
This means that 2870 kJ is the maximum amount of work that can be done by a person as a result of metabolizing 1 mol of glucose. Suppose a man (mass 60 kg) climbs a height of 100 m . The work a person must do in climbing a height h is given by , w = m g h where g is accleration due to gravity (9.8 m s-2). Thus the man has to work
Work has to do = (60 kg) (9.8 m s-2) x 100 m
= 58,800 kg m2 s-2 = 58,800 J = 58.8 kJ
In order to do this much work, he need to metabolize minimum of (58.8 / 2870) = 0.0205 mol or 0.0205 x 180 = 3.69 g of glucose.
However, in practice, the conversion of energy to the work is not 100% efficient , therefore the person would need more than 3.7 g of glucose to climb the height.
Gibb’s Energy Change and Electrical work done
In the case of electrochemical cells, Gibb’s energy change DG is related to the electrical work done in the cell. If E is the EMF of the cell and n mole of electrons are involved , the electrical work will be :
DG = - n F E
where F is Faraday constant = 96,500 C
If the products are in their standard states :
DGΘ = - n F EΘ
where EΘ is the standard cell potential.
Gibb’s Energy Changes for predicting feasibility of a reaction
Let us develop the criterion of spontaneity in terms of Gibb’s energy change. The total entropy change during the process is given by :
DStotal = DSsystem + DSsurrounding …… (3)
Consider a process being carried out at constant temperature and pressure. Suppose heat equal to q is lost by the surroundings. Now heat gained by the system at constant temperature and pressure qp represents its enthalpy change (DH). Therefore.
Negative sign before q indicates that total heat has been lost by the surroundings. Now the heat gained by the system at constant temperature and pressure qp represents the enthalpy change (DH). Therefore,
Substituting this value in equation (12) , we get :
Multiplying throughout by T :
The equation (5) forms the basis for the criterion of spontaneity in terms of Gibb’s energy change of the system only. For example, for a spontaneous process DStotal is positive. So that DGsystem is negative.
Thus for a process, at a particular temperature and pressure to be spontaneous , the Gibb’s energy of the system must decrease, i.e., DGT,P < 0.
Summary of Gibb’s energy criteria of spontaneity
i) If DStotal is positive : DG.T,P is negative. The process will be spontaneous.
ii) If DStotal = 0 ; DGT,P = 0 ; the process will be at equilibrium.
iii) If DStotal is negative : DGT,P is positive. The process will
be non-spontaneous.
Conditions for DG to be negative
Let us formulate the conditions for DG to be negative. It can be done by considering Gibb’s Helmholtz equation.
DG = DH - T DS
i) When both energy factor as well as entropy factor are favourable, i.e., DH is negative and DS is positive. Under these conditions DG would certainly be negative and the process would occur spontaneously.
ii) Energy factor favours but entropy factor opposes, i.e., DH is negative and DS is negative. Under these conditions DG would be negative if DH is greater than TDS in magnitude.
iii) Energy factor opposes but entropy factor favours, i.e., DH is positive and DS is also positive. Under these conditions, DG would be negative if TDS is greater than DH in magnitude.
These conditions have been summarised in the following TABLE.
Sign of DH | Sign of DS | Sign of DG, occurrence of a reaction |
- | + | - ve (spontaneous) |
- | - | - ve ( if DH > TDS) |
+ | + | - ve ( if TDS > DH) |
Effect of Temperature on Feasibility of a Reaction
Gibb’s Helmholtz equation helps us in predicting the feasibility of a process at a particular temperature. The spontaneity of a process depends on :
i) the energy factor DH and
ii) the entropy factor DS.
As is evident from the equation : DG = DH - T DS
the term TDS may assume large values at higher temperatures due to increase in multiplying factor T as well as increase in randomness. But the other factor DH does not vary appreciably with temperature. Therefore, the free energy change (overall tendency) of a process to take place is also influenced by temperature. The effect of temperature is different for exothermic and endothermic reactions as discussed below:
a) Exothermic Reactions
For exothermic reactions, DH is always negative and therefore , it is favourable. Now TDS can either positive or negative value.
i) If TDS is positive, i.e., favourable, then DG can have only negative value and the process is spontaneous at all temperatures.
ii) If TDS is also negative, i.e., unfavourable. The value of DG will be negative only if DH > TDS. This fact will be more predominant at lower temperatures which makes the contributions of TDS(opposing factor) less. Hence exothermic reactions are favoured by decreasing temperature.
b) Endothermic reactions
For endothermic reactions, DH is positive and always opposes the process. Now :
i) If TDS is negative and opposes the process , then DG will be positive and the process is always non-spontaneous.
ii) On the other hand, when TDS is positive (favourable) , the value of DG will be negative only if TDS > DH. This fact will be more predominant at higher temperatures which makes the contribution of TDS (favouring factor) more. This means that endothermic reactions are favoured by increasing temperature. These results are summed in the following TABLE.
TABLE
Tendencies of reactions to occur in forward direction and effect of temperature.
Reaction | DH | Sign of TDS | DG | Behaviour |
Exothermic | - | + | - | spontaneous |
- | - | - at low T + at high T | spontaneous Non-spontaneous | |
Endother -mic | + | - | + | non-spontaneous |
+ | + | + at low T | non-spontaneous | |
+ | + | - at high T | spontaneous |
Problems
69. Consider the reaction for the dissolution of ammonium nitrate
NH4NO3(s) ® NH4+(aq) + NO3-(aq)
DH = +28.1 kJ/mol : DS = 108.7 kJ mol-1
Calculate the change in entropy of the surroundings and predict whether the reaction is spontaneous or not at 298 K ?
70. The rusting of iron occurs as :
4 Fe(s) + 3 O2(g) ® 2 Fe2O3
The enthalpy of formation of Fe2O3(s) is - 824.2 kJ mol-1 and entropy change for the reaction is -549 J K-1 . Calculate DSsur and predict whether rusting of iron is spontaneous or not at 298 K.
71. For the reaction :
2 H2O2(g) ® 2 H2O(g) + O2(g)
DH = - 211.29 kJ/mol and DS = + 131.98 J K-1 mol-1.
Predict the feasibility of the reaction at 298 K.
72. For the reaction :
N2(g) + O2(g) ® 2 NO(g)
DH = +180.5 kJ : DS = 24.6 J K-1 at 298 K.
CalculateDG.
73. For the melting of ice at 298 K ,
H2O (s) ® 2 H2O(ℓ)
The enthalpy of fusion is 6.97 kJ mol-1 and entropy of fusion is 25.4 J mol-1. Calculate the free energy change and predict whether melting of ice is spontaneous at this temperature.
74. Enthalpy and entropy changes of a reaction are 40.63 kJ/mol and 108.8 kJ/mol respectively. Predict the feasibility of the reaction at 300 K.
75. DH and DS for the reaction :
Ag2O(s) ® 2 Ag(s) + ½ O2(g)
are 30.56 3 kJ mol-1and 108.8 J K-1mol-1 respectively. Predict the feasibility of the reaction at 300 K.
Standard Gibb’s energy change (DGΘ)
The standard Gibb’s energy change is the free energy for a process at 298 K and one atmospheric pressure in which the reactants in their standard states are converted to the products in their standard states. It is denoted by DGΘ . The standard Gibb’s energy change is related to the standard ethalpy change by the relation :
DGo= DHΘ - T DSΘ
where DHo and DSo represent the standard enthalpy change and standard entropy change during the process respectively.
Like DHo the standard free energy change (DGΘ) of a reaction can be calculated from the standard Gibb’s energy of formation of the products and that of the reactants. The standard free energy of formation (DfGΘ ) is defined as the Gibb’s energy change taking place when 1 mole of the compound is formed from the constituent elements in the standard states. It may be noted that the standard free energy of formation (DfGΘ ) of elementary substances in their most stable form is taken to be zero.
The standard free energy of a reaction can be calculated from the knowledge of standard free energy of formation of various substances as follows :
DG0 = SDfGo (products) - SDfGo (reactants)
= [Sum of standard free energies of formation of products]
- [Sum of standard free energies of formation reactants]
For example, for a reaction :
a A + b B ® ℓ L + m M
The standard free energy change can be calculated as :
DGo= SDfGΘ (products) - SDfGΘ (reactants)
= [ℓ DfGΘ(L) + mDfGΘ (M)] -[a DfGΘ (A) + b DfGΘ (B)]
COUPLED REACTIONS
There are reactions for which the value of DrG is not negative i.e., they are non-spontaneous. However, such reactions can be made spontaneous if these are coupled (means carrying both reactions simultaneously) with reactions having very large negative free energy of reaction.
Let us consider the decomposition of iron oxide , Fe2O3 into iron :
2 Fe2O3(s) ® 4 Fe(s) + 3 O2(g)
: DGΘ =+1487.0 kJ mol-1 (a)
The positive DG shows that the reaction is non-spontaneous. This can be achieved by coupling decomposition of Fe2O3 with a reaction having a highly negative DrGΘ .
Let us consider the reaction :
2 CO(g) + O2(g) ® 6 CO2 (g)
: DrGΘ = -514.4 kJ mol-1……(b)
In Equ. (a) 3 mol of O2 is involved. Therefore , we can write equation (b) as
6 CO(g) + 3O2(g) ® 6 CO2(g)
: DrGΘ = -1543.2 kJ mol-1 ……(c)
Adding equations (a) and (c)
2 Fe2O3(s) ® 4 Fe(s) + 3 O2(g) : DGΘ =+1487.0 kJ mol-1
6 CO(g) + 3O2(g) ® 6 CO2(g) : DrGΘ = -1543.2 kJ mol-1
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2 Fe2O3(s) + 6 CO(g) ® 4 Fe(s) +6 CO2(g)
: DGΘ = -56.2 kJ mol-1
The negative value of DGΘ indicates that iron(III) oxide can be reduced spontaneously to free iron and carbon dioxide. This is reaction which takes place in the lower part of blast furnace where iron ore is reduced to iron.
The concept of coupling two reactions (one spontaneous and one non-spontaneous) is very useful in biological systems. Adenosine triphosphate , or ATP , is a large molecule containing phosphate groups. When ATP reacts with water in presence of an enzyme, it is hydrolysed to give adenosine diphosphate, ADP , and a phosphate ion.
ATP + H2O ® ADP + phosphate ion : DrGΘ = - 31 kJ mol-1
This reaction is coupled with various non-spontaneous reactions to carry out the necessary reactions in biological systems. For example, the biosynthesis of sucrose from glucose and fructose has DGΘ of 23 3 kJ mol-1. Hence , this reaction can be driven by hydrolysis of ATP.
Glucose + Fructose + ATP ® Sucrose + ADP : DGΘ = - 8 kJ/mol
Therefore ATP is regarded as centre of activities of the cell.
Problem
76. Calculate the standard Gibb’senergy change for the formation of methane at 298 K.
C(graphite) + 2 H2(g) ® CH4(g)
DfHΘCH4(g) = - 74.81 J mol-1. SΘ J K-1 mol-1 : C(graphite) = 5.70 : H2(g) = 130.7 : CH4(g) = 186.3.
77. Silane SiH4 burns in air . The product silica is a solid.
SiH4(g) + 2 O2(g) ® SiO2(s) + 2 H2O(g)
Standard Gibb’s energy of formation of SiH4(g) is +52.3 kJ mol-1. Values for SiO2 and H2O(g) are 805 and -228.6 kJ mol-1 respectively. Calculate DrGΘ .
60. Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature it becomes spontaneous. For the reaction
MgO(s) + C(s) ® Mg(s) + CO(g) : DrHΘ= + 491.18 kJ mol-1 and DrSΘ = 197.67 J K-1 mol-1.
78. In a fuel cell , methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is :
CH3OH(ℓ) + 3/2 O2(g) ® CO2(g) + 2 H2O(ℓ)
Calculate the standard Gibb’s energy change for the reaction that can be converted into electrical work. If standard enthalpy change of combustion for methanol is -726 kJ/mol, calculate the efficiency of conversion of Gibb’s energy into useful work.
Gibb’s Energy Change and Equilibrium Constant
For chemical reactions to occur, Gibb’s energy change should be negative. During the course of the reaction, the Gibb’s energy decreases and composition of reaction mixture changes with time and finally a constat composition of mixture is obtained. A curve of the shape (shown in Fig) is obtained.
At constant temperature and pressure, the direction of spontaneous change is towards lower Gibb’s energy. The equilibrium composition of a reaction mixture corresponds to the lowest point on the curve. In this example substatial quantities of both reactants and products are present at equilibrium and K is close to 1.
The composition at lowest point of the curve – point of minimum Gibb’s energy – corresponds to equilibrium. For a system at equilibrium , any change either in the forward reaction or reverse reaction, would lead to increase in Gibb’s energy. In other words, when system attains equilibrium, any change in forward or reverse direction is not spontaneous.
The following generalisations can be made from the above diagram :
· When Gibb’s energy minimum lies very close to the products, the equilibrium composition strongly favours products. In other words, the reaction goes nearly to completion i.e., K >> 1. This is shown in Fig (a)
(a) At equilibrium , the products are much more in abundance than the reactants ( K >> 1).
· When the Gibb’s energy minimum lies very close to the reactants, the equilibrium composition favours the reactants and the reaction forms only a small amounts of products. In other words, the reaction does not proceed much i.e., K << 1. This is shown in Fig (b).
(b) Equilibrium lies in favour of reactants ( K << 1).
· When the Gibb’s energy minimum lies approximately half-way between the reactants and products, both the reactants and products are present in almost equal concentration at equilibrium , i.e., K is close to 1.
In order to make these ideas quantitative, we must consider Gibb’s energy change , DrG and standard Gibb’s energy change DrGΘ . Here DrGΘ is the difference in standard Gibb’s energies of products and reactants both in their standard states. Similarly, DrG is the reaction Gibb’s energy change at a definite, fixed composition of reaction mixture.
The Gibb’s energy of reaction DrG , is related to composition of the reaction mixture and to standard reaction Gibb’s energy by the equation :
DrG =DrGΘ + R T ln Q
where Q is the reaction quotient and R is the gas constant, with the value 8.314 J/K/mol . If the species are gases, these concentrations are expressed in partial pressure and the reaction quotient will be Qp and if the species are in solution, reaction quotient will be expressed in terms of molar concentration as Qc.
At equilibrium , Q = K and therefore, above equation becomes :
0 = DrGΘ + R T ln K
or DrGΘ = -R T ln K
or DrGΘ = - 2.303 R T log K …. (1)
The equation can also be written as :
We also know that
DrG = DrHΘ - T DS = - R T ln K
Since DrGΘ depends on the value of DrHΘ, for strongly endothermic reactions the value of DrHΘ may be large and positive. In such case the value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, DrHΘ is large and negative and DrGΘ is likely to be large and negative too. In such cases , K will be much larger than 1. We may expect strongly exothermic reactions to have a large K , and hence can go to near completion.
Using equation (1) we can determine the value of DrGΘ if the value of K is known or vice versa.
Problems
79. For the water gas reaction:
C(s) + H2O(g) ⇌CO(g) + H2(g)
The standard Gibb’s energy of reaction (at 1000 K) is - 8.1 kJ mol-1. Calculate equilibrium constant.
80. The standard Gibb’s energies for the reaction at 1773 K are given below :
C(s) + O2(g) ® CO2(g) : DrG⊝= - 380 kJ mol-1
2 C + O2(g) ⇌ 2 CO2(g) : DrG⊝= - 500 kJ mol-1
Discuss the possibility of reducing Al2O3 and PbO with carbon at this temperature.
81. Consider the following reaction :
2 NO(g) + O2(g) ® 2 NO2(g)
Calculate the value of standard Gibb’s energy at 298 K and predict whether the reaction is spontaneous or not.
82. It is planned to carry the reaction :
CaCO3(s) ⇌ CaO(s) + CO2(g)
a 1273 K at 1 bar.
(a) Is this reaction spontaneous at this temperature and pressure.
(b) Calculate the value of :
(i) Kp at 1273 K for the reaction.
(ii) Partial pressure of CO2 at equilibrium.
83. The equilibrium constant for the reaction :
PCl5 (g) ⇌ PCl3(g) + Cl2(g)
at 298 K has been found to be 1.8 x 10-7. Calculate DGΘ for the reaction.
84. Using the following data , calculate the value of equilibrium constant for the following reaction at 298 K.
2 CHºCH ⇌ C6H6(g)
assuming ideal behaviour
DfGΘ (CHºCH) = 2.09 x 105 J/mol
DfGΘ (C6H6) = 1.24 x 105 J/mol
Can the reaction be recommended for the synthesis of benzene ?
85. Calculate equilibrium constant for the reaction at 400 K.
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g) : DHΘ = 77.2 kJ/mol ;
DS= 122 J K-1mol-1 at 400 K.
86. The equilibrium constant at 298 K for the process :
Co3+(aq) + 6 NH3 ⇌ [Co(NH3)6]3+(aq)
is 2.0 x 107 . Calculate the value of DGΘ at 298 K.
87. The DrGΘ for conversion of oxygen to ozone
3/2 O2(g) ® O3(g)
at 298 K , if Kp for this conversion is 2.47 x 10-29.
88. Find out the value of equilibrium constant for the following reaction at 298 K.
2 NH3(g) + CO2(g) ⇌ NH2CONH2(aq) + H2O(ℓ)
Standard Gibb’s energy change, DrGΘ , at the given temperature is -13.6 kJ mol-1.
89. Calculate standard free energy change for the reaction :
Zn + Cu2+(aq) ® Cu + Zn2+(aq) : EΘ = 1.10 V
90. Calculate at 298 K , the equilibrium constant for the reaction :
Cu(s) + 2 Ag+(aq) ® Cu2+(aq) + 2 Ag(s)
At 298 K EΘ cell = 0.47 V.
Questions
1. What is meant by the term 'chemical energetics'?
2. What are primary sources of energy ?
3. What is meant by renewable and non-renewable source of energy ? Give examples.
4. In view of the limited availability of non-renewable sources suggest suitable energy sources for future needs of the country.
5. Explain the terms : i) Hydro-power and ii) Geothermal power.
6. Explain the term chemical energy.
7. Explain the term 'internal energy'.
8. Explain the term 'internal energy change'.
9. What is meant by term 'state property'?
10. Explain the terms : I) Thermodynamic system and ii) Surroundings.
11. What the origin of energy change in a chemical reaction ?
12. Define the terms : I) Open system ii) Closed system iii) isolated system iv) extensive property v) intensive property vi) isothermal process vii) adiabatic process.
13. State and explain the Law of Conservation of energy.
14. What are spontaneous reactions ? Give suitable examples.
15. Define Enthalpy of a system.
16. How is enthalpy related to the internal energy of the system ?
17. Mention the factors which influence the value of enthalpy change in a chemical reaction. Define standard molar enthalpy change.
18. Define exothermic and endothermic reactions.
19. State and explain Hess's Law of constant heat summation by taking suitable examples.
20. Write a mathematical relationship between heat, internal energy and work done by the system.
21. Explain the term heat of reaction. Give two examples.
22. Explain the terms : I) Enthalpy of combustion. ii) Enthalpy of formation iii) Enthalpy of vaporisation iv) Enthalpy of fusion v) Enthalpy of sublimation
23. What is bond energy ?
24. Explain why the enthalpy of neutralisation of weak acid by a strong base is less than that of strong acid ?
25. The heat of combustion of methane is - 890.65 kJ/mol and heats of formation of CO2 and H2O are - 393.5 and - 286.0 kJ/mol respectively. Calculate the heat of formation of methane (ans : - 74.85 kJ)
26. Write notes on :
(i) entropy of a system
(ii) free energy of a chemical reaction
27. Why the change in enthalpy cannot be a sole criterion for the spontaneity of a process ?
28. In relation to DH, DS and DG values for a reaction, when could it :
(i) proceed forward spontaneously
(ii) be at equilibrium
29. Why does entropy of a solid increases on fusion ?
30. Why is entropy of ice less than that of water ?
31. Indicate whether the entropy increases or decreases in the following changes. Give reasons :
(i) evaporation of water
(ii) CaCO3(s) ® CaO(s) + CO2(g)
(iii) 2 SO2(g) + O2(g) ® 2 SO3(g)
32. Discuss the nature of change in entropy(increase or decrease) for the following :
(i) H2O(l) ® H2O(s)
(ii) MgCO3(s) ® MgO(s) + CO2(g)
(iii) 2 SO2(g) + O2(g) ® 2 SO3(g)
33. State the relationship amongst changes in free energy, enthalpy and entropy of a system at a given temperature. How does these changes , being small or large or negative of positive , affect the feasibility of a process ?
34. Predict the feasibility of a reaction when :
(a) both DH and DS increase ;
(b) both DH and DS decrease
(c) DH decreases, DS increases.
35. Discuss the conditions for spontaneous occurrence of a process.
36. What is meant by a ‘fuel’ ?
37. What is meant by calorific value of a fuel ?
38. Write a note on calorific value of foods.
39. Will the temperature of the surrounding air increase or decrease, if :
(a) an exothermic reaction is allowed to occur very rapidly in air ?
(b) an endothermic reaction is allowed to occur ver rapidly in air ?
40. Discuss the significance of the mathematical expression in which the heat absorbed by a system is related to internal energy and work done by the system.
41. How Gibb’s free energy is related to enthalpy, entropy and temperature of a system ? How is this used in determining the spontaneity of a process ?
42. Why is bond energy of Cl - Cl less than that of H- H ?
43. Predict if the following reaction :
CH3CH2OH(g) ® CH2=CH2(g) + H2O(g)
will be exothermic or endothermic. Give reason for
your answer.
44. Why is more energy required to break C- H bond in acetylene compared to ethane ?
45. Why is bond energy of ethene is less than that of acetylene ?
46. Explain the following terms :
(a) zeroth law of thermodynamics.
(b) Reversible and irreversible process
47. Explain two alternative energy sources for our future.
48. Define heat capacity, specific heat and molar heat capacity. How are they related ?
49. Define the terms:
(i) Enthalpy of fusion.
(ii) Enthalpy of vaporisation.
(iii) Enthalpy of sublimation
How are these terms related ?
50. What are extensive and intensive properties ?
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